Problem: The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives $14$ years; the standard deviation is $1.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living between $16.6$ and $17.9$ years.
$14$ $12.7$ $15.3$ $11.4$ $16.6$ $10.1$ $17.9$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $14$ years. We know the standard deviation is $1.3$ years, so one standard deviation below the mean is $12.7$ years and one standard deviation above the mean is $15.3$ years. Two standard deviations below the mean is $11.4$ years and two standard deviations above the mean is $16.6$ years. Three standard deviations below the mean is $10.1$ years and three standard deviations above the mean is $17.9$ years. We are interested in the probability of a meerkat living between $16.6$ and $17.9$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the meerkats will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the meerkats will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of meerkats between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular meerkat living between $16.6$ and $17.9$ years is $\color{orange}{2.35\%}$.